Chapter 4: Clustering and classification
Anni Norring
23 marraskuuta 2017
Chapter 4: Clustering and classification
RStudio exercise 4: Data analysis and wrangling
Data analysis
2. Data
The Boston dataset contains data from a study on housing values in suburbs of Boston. With the data you can for example study the effect of crime rates or air quality on the value of owner-ocupied homes. You can learn more about the data here.
The dataset contains 506 observations on 14 variables. The observations are numeric with most variables taking natural number values. There are also some with integer values and one dichotomous variable.
# load the data
data("Boston")
# glimpse at the dataset
glimpse(Boston)## Observations: 506
## Variables: 14
## $ crim <dbl> 0.00632, 0.02731, 0.02729, 0.03237, 0.06905, 0.02985, ...
## $ zn <dbl> 18.0, 0.0, 0.0, 0.0, 0.0, 0.0, 12.5, 12.5, 12.5, 12.5,...
## $ indus <dbl> 2.31, 7.07, 7.07, 2.18, 2.18, 2.18, 7.87, 7.87, 7.87, ...
## $ chas <int> 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
## $ nox <dbl> 0.538, 0.469, 0.469, 0.458, 0.458, 0.458, 0.524, 0.524...
## $ rm <dbl> 6.575, 6.421, 7.185, 6.998, 7.147, 6.430, 6.012, 6.172...
## $ age <dbl> 65.2, 78.9, 61.1, 45.8, 54.2, 58.7, 66.6, 96.1, 100.0,...
## $ dis <dbl> 4.0900, 4.9671, 4.9671, 6.0622, 6.0622, 6.0622, 5.5605...
## $ rad <int> 1, 2, 2, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, ...
## $ tax <dbl> 296, 242, 242, 222, 222, 222, 311, 311, 311, 311, 311,...
## $ ptratio <dbl> 15.3, 17.8, 17.8, 18.7, 18.7, 18.7, 15.2, 15.2, 15.2, ...
## $ black <dbl> 396.90, 396.90, 392.83, 394.63, 396.90, 394.12, 395.60...
## $ lstat <dbl> 4.98, 9.14, 4.03, 2.94, 5.33, 5.21, 12.43, 19.15, 29.9...
## $ medv <dbl> 24.0, 21.6, 34.7, 33.4, 36.2, 28.7, 22.9, 27.1, 16.5, ...3. Overview of the data
Next we will look at the data in a bit more detail. Start by printing out the column/variable names and writing down the definitions of the different variables (from here):
colnames(Boston)## [1] "crim" "zn" "indus" "chas" "nox" "rm" "age"
## [8] "dis" "rad" "tax" "ptratio" "black" "lstat" "medv"| Variable | Definition |
|---|---|
| crim | per capita crime rate by town |
| zn | proportion of residential land zoned for lots over 25,000 sq.ft |
| indus | proportion of non-retail business acres per town |
| chas | Charles River dummy variable (= 1 if tract bounds river; 0 otherwise) |
| nox | nitrogen oxides concentration (parts per 10 million) |
| rm | average number of rooms per dwelling |
| age | proportion of owner-occupied units built prior to 1940 |
| dis | weighted mean of distances to five Boston employment centres |
| rad | index of accessibility to radial highways |
| tax | full-value property-tax rate per $10,000 |
| ptratio | pupil-teacher ratio by town |
| black | 1000(Bk - 0.63)^2 where Bk is the proportion of blacks by town |
| lstat | lower status of the population (percent) |
| medv | median value of owner-occupied homes in $1000s |
From the summary of the variables we can see minimum, maximum, median and mean values as well as the 1st and 3rd quartiles of the variables.
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00We can study and visualize the correlations between the different variables with the help of a correlations matrix and a correlations plot.
# First calculate the correlation matrix and round it so that it includes only two digits:
cor_matrix<-cor(Boston) %>% round(digits = 2)
# Print the correlation matrix:
cor_matrix## crim zn indus chas nox rm age dis rad tax
## crim 1.00 -0.20 0.41 -0.06 0.42 -0.22 0.35 -0.38 0.63 0.58
## zn -0.20 1.00 -0.53 -0.04 -0.52 0.31 -0.57 0.66 -0.31 -0.31
## indus 0.41 -0.53 1.00 0.06 0.76 -0.39 0.64 -0.71 0.60 0.72
## chas -0.06 -0.04 0.06 1.00 0.09 0.09 0.09 -0.10 -0.01 -0.04
## nox 0.42 -0.52 0.76 0.09 1.00 -0.30 0.73 -0.77 0.61 0.67
## rm -0.22 0.31 -0.39 0.09 -0.30 1.00 -0.24 0.21 -0.21 -0.29
## age 0.35 -0.57 0.64 0.09 0.73 -0.24 1.00 -0.75 0.46 0.51
## dis -0.38 0.66 -0.71 -0.10 -0.77 0.21 -0.75 1.00 -0.49 -0.53
## rad 0.63 -0.31 0.60 -0.01 0.61 -0.21 0.46 -0.49 1.00 0.91
## tax 0.58 -0.31 0.72 -0.04 0.67 -0.29 0.51 -0.53 0.91 1.00
## ptratio 0.29 -0.39 0.38 -0.12 0.19 -0.36 0.26 -0.23 0.46 0.46
## black -0.39 0.18 -0.36 0.05 -0.38 0.13 -0.27 0.29 -0.44 -0.44
## lstat 0.46 -0.41 0.60 -0.05 0.59 -0.61 0.60 -0.50 0.49 0.54
## medv -0.39 0.36 -0.48 0.18 -0.43 0.70 -0.38 0.25 -0.38 -0.47
## ptratio black lstat medv
## crim 0.29 -0.39 0.46 -0.39
## zn -0.39 0.18 -0.41 0.36
## indus 0.38 -0.36 0.60 -0.48
## chas -0.12 0.05 -0.05 0.18
## nox 0.19 -0.38 0.59 -0.43
## rm -0.36 0.13 -0.61 0.70
## age 0.26 -0.27 0.60 -0.38
## dis -0.23 0.29 -0.50 0.25
## rad 0.46 -0.44 0.49 -0.38
## tax 0.46 -0.44 0.54 -0.47
## ptratio 1.00 -0.18 0.37 -0.51
## black -0.18 1.00 -0.37 0.33
## lstat 0.37 -0.37 1.00 -0.74
## medv -0.51 0.33 -0.74 1.00# Visualize the correlation matrix with a correlations plot:
corrplot(cor_matrix, method="circle", type = "upper", cl.pos = "b", tl.pos = "d", tl.cex = 0.6)
From the plot above we can easily see which variables correlate with which and is that correlation positive or negative. So for example, we can see that the distance from the employment centres (dis) correlates negatively with age of the houses (age), nitrogen oxides concentration (nox) and proportion of non-retail business acres (indus). We can also see that accessibility to radial highways (rad) is highly positively correlated with property taxes (tax), as are industrial land use (indus) and air pollution (nox). On the first glimpse, the correlation appear quite intuitive and nothing stands out as very suprising. One could note that the correlation of the only dummy variable chas is almost negligible with all other variables. This could be because the overall share of observations where chas = 1 is very small.
4. Standardizing and scaling the dataset
We will later use linear discriminant analysis on our data, so we need to have varibles that are normally distributed and share the same covariance matrix across different classes. This requires the data to be scaled before fitting a model.
Thus we want to standardize the data set by scaling it. This is very easy to do in R by using the scale() function, which subtracts the mean of a variable from each value of the variable and divides this by the standard error. From the summary we can see that the standardized values are much more similar in magnitude across the different variables, so their relationships are easier to grasp. Note that the mean of each variable is now zero, as it should be. Also note that previously all variables were positive, but now at least the min values are negative for all variables. This is because the mean of all variables is larger than the smallest value of a variable.
# center and standardize variables
boston_scaled <- scale(Boston)
# summaries of the scaled variables
summary(boston_scaled)## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865Note that the scaled data set is now in matrix format. We want to change the class of the scaled data set into a dataframe, which we can do by converting it with a function as.data.frame():
# class of the boston_scaled object
class(boston_scaled)## [1] "matrix"# change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)Next we want to create a categorical (or factor) variable of the crime rate in the scaled Boston data set. We will use the quantiles as the break points in the categorical variable. We will print the summary of the scaled crime rate, the quantile vector of this variable and a table of the new factor variable to check that everything goes as intended.
# summary of the scaled crime rate
summary(boston_scaled$crim)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -0.419367 -0.410563 -0.390280 0.000000 0.007389 9.924110# create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
bins## 0% 25% 50% 75% 100%
## -0.419366929 -0.410563278 -0.390280295 0.007389247 9.924109610# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, label = c("low", "med_low", "med_high", "high"))
# look at the table of the new factor crime
table(crime)## crime
## low med_low med_high high
## 127 126 126 127In order to avoid confusion, we want to drop the old crime rate variable from the dataset and replace it with the new categorical variable for crime rates. This is easily done with the following two lines of code:
# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)
# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)Lastly for this step, we will divide the dataset to train and test sets. This will allow as further down the line to check how well our model works. We will assign 80 % of the data to the train set and the remaining 20 % to the test set. The training of the model is done with the train set and predictions on new data is conducted on the test set.
# number of rows in the Boston dataset
n <- nrow(boston_scaled)
# choose randomly 80% of the rows
ind <- sample(n, size = n * 0.8)
# create train set
train <- boston_scaled[ind,]
# create test set
test <- boston_scaled[-ind,]
# save the correct classes from test data
correct_classes <- test$crime
# remove the crime variable from test data
test <- dplyr::select(test, -crime)5. Linear discriminant analysis
Now we will fit the linear discriminant analysis on the train set. The LDA is a classification method that finds the (linear) combination of the variables that separate the target variable classes. We will use the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables.
# linear discriminant analysis
lda.fit <- lda(crime ~., data = train)
# print the lda.fit object
lda.fit## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.2524752 0.2400990 0.2574257 0.2500000
##
## Group means:
## zn indus chas nox rm
## low 0.95313734 -0.8920844 -0.0793339581 -0.8664560 0.43807213
## med_low -0.06509931 -0.2951030 0.0117915679 -0.5979929 -0.11185772
## med_high -0.40395960 0.2353055 0.1819517287 0.3866572 -0.06085731
## high -0.48724019 1.0149946 0.0005392655 1.0522191 -0.42284726
## age dis rad tax ptratio
## low -0.9062951 0.8977263 -0.6868728 -0.7373246 -0.41147812
## med_low -0.3844676 0.3848914 -0.5615559 -0.5305818 -0.05707988
## med_high 0.4165048 -0.3824748 -0.4164721 -0.2947116 -0.25216256
## high 0.8359851 -0.8645371 1.6596029 1.5294129 0.80577843
## black lstat medv
## low 0.3792107 -0.7693183 0.534112876
## med_low 0.3524186 -0.1727156 0.007978516
## med_high 0.1023849 0.0866083 0.068152872
## high -0.7900314 0.8681775 -0.662513146
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.156321053 0.676225751 -0.8905832
## indus -0.019547429 -0.200159101 0.4065107
## chas -0.001110237 -0.045017648 0.1011732
## nox 0.274713478 -0.641845168 -1.4216678
## rm 0.027025317 0.002053063 -0.1407244
## age 0.333763752 -0.320024481 -0.1078748
## dis -0.191770489 -0.179940972 0.1811128
## rad 3.400254119 1.058560042 0.1862033
## tax 0.033258236 -0.190826857 0.3077754
## ptratio 0.169242647 0.080546275 -0.2537049
## black -0.171057921 0.018548903 0.1741080
## lstat 0.106331768 -0.213377185 0.2439645
## medv 0.030763748 -0.309750086 -0.3225459
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9577 0.0325 0.0099The LDA calculates the probabilities of a new observation belonging to each of the classes by using the trained model. After that the model classifies each observation to the most probable class.
The most convenient and informative way to consider the results of an LDA is to visualize with a biplot. This we can do following the Datacamp exercise with the following code:
# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 2)
The arrows in this plot represent the relationships of the original variables and the LDA solution.
6. LDA predicting performance
In this part we want to see how the LDA model performs when predicting on new data. For this end, we can use the predict() function. we start by predicting the crime classes with the test data.
# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)
# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)## predicted
## correct low med_low med_high high
## low 16 9 0 0
## med_low 2 17 10 0
## med_high 0 4 17 1
## high 0 0 1 25From the cross tabulation of the results we can see that the model appears to predict the correct results reasonably well. The model has most problems in separating med_low from low, but most of the predictions seems to fall in the correct class.
7. Clustering
Next we will take few steps towards clustering the data. Start by reloading the Boston dataset and standardize the dataset again. We need to do this again, because we have made also other changes to dataset.
# load the data
data("Boston")
# center and standardize variables
boston_scaled2 <- scale(Boston)
# summaries of the scaled variables
summary(boston_scaled2)## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865# class of the boston_scaled object
class(boston_scaled2)## [1] "matrix"# change the object to data frame
boston_scaled2 <- as.data.frame(boston_scaled)Next we wish to calculate the distances between the observations. We do this by forming a standard Euclidian distance matrix:
# euclidean distance matrix
dist_eu <- dist(boston_scaled2)## Warning in dist(boston_scaled2): NAs introduced by coercion# look at the summary of the distances
summary(dist_eu)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1394 3.5267 4.9081 4.9394 6.2421 13.0045I ran into trouble here because there are NA values in the boston_scaled2 distance matrix. The k-means algorithm wouldn’t run on this dataset, retuning an error on the NA values. Despite googling I was not able to find a solution, so I’m doing the rest of this exercise with the original Boston dataset (as in the Datacamp exercises). So, a novel try:
# euclidean distance matrix
dist_eu <- dist(Boston)
# look at the summary of the distances
summary(dist_eu)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.119 85.624 170.539 226.315 371.950 626.047From the summary we can see the relevant moments of the calculated distances between the observations.
Then let’s move on to K-means clustering, that assigns observations to groups or clusters based on similarity of the objects. The following code runs k-means algorithm on the dataset:
# k-means clustering
km <-kmeans(Boston, centers = 3)
# plot the Boston dataset with clusters
pairs(Boston, col = km$cluster)
It’s difficult to see anything from this, so let’s zoom in a bit and look at only the last five columns:
# plot the Boston dataset with clusters
pairs(Boston[6:10], col = km$cluster)
Next we want to know what the optimal number of clusters is. We can do this by looking at how the total of within cluster sum of squares (WCSS) behaves when the number of cluster changes. When you plot the number of clusters and the total WCSS, the optimal number of clusters is when the total WCSS drops radically.
# MASS, ggplot2 and Boston dataset are available
set.seed(123)
# determine the number of clusters
k_max <- 10
# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(Boston, k)$tot.withinss})
# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')
From the plot above it’s evident that optimal number of clusters is not three, but two. Let’s run the k-means algorithm again with two clusters and then visualize the results:
# k-means clustering
km <-kmeans(Boston, centers = 2)
# plot the Boston dataset with clusters
pairs(Boston, col = km$cluster)
# plot the Boston dataset with clusters
pairs(Boston[6:10], col = km$cluster)
Super-bonus
Start by running the code below for the (scaled) train data that we used to fit the LDA. The code creates a matrix product, which is a projection of the data points.
model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)## Warning: package 'plotly' was built under R version 3.4.2##
## Attaching package: 'plotly'## The following object is masked from 'package:MASS':
##
## select## The following object is masked from 'package:ggplot2':
##
## last_plot## The following object is masked from 'package:stats':
##
## filter## The following object is masked from 'package:graphics':
##
## layoutThen we will use the package plotly to create a 3D plot of the columns of the matrix product:
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')## Warning: package 'bindrcpp' was built under R version 3.4.2This is very cool indeed! But addingdifferent colors for the different crime rate classes definetly makes it even cooler still:
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = train$crime)plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = km)Data wrangling
You can find my preprocessed data set and the data wrangling R script from my GitHub repository.